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Mendelian Genetics

Learn the foundational concepts of Mendelian genetics, including terms such as gene, allele, trait, genotype, phenotype, and more. Explore Mendel's famous experiments and the laws of segregation and independent assortment. Understand dominance, incomplete dominance, co-dominance, and lethal alleles. Study gene-gene interactions, including epistasis and collaborative and supplementary gene interactions.

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Mendelian Genetics

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  1. Mendelian Genetics Read through core knowledge. What vocab do you need to learn?

  2. Terms • Gene • Allele • Trait • Dominant • Recessive • Homozygous • Heterozygous • Genotype • Phenotype • P and F1 and F2

  3. Definitions • Unit of hereditary • One of 2 or more forms of a gene at a given locus • Genetically inherited characteristic of organism, varies amongst individuals • Allele that is expressed in heterozygotes • Allele that is only expressed in homozygotes • Carries two copies of the allele • Carries different allelic forms of a given gene • Organism’s hereditary make-up • Physical characteristics of an organism • Patrial generation, first and second filial generation

  4. Mendel – why so famous? • Worked with peas • Used pure-breeding varieties • Came up with idea of ‘gene’ 20 years before chromosomes were discovered

  5. Law of segregation • Of the two genes controlling each characteristic, only one is present in each gamete. • During meiosis the two genes are separated.

  6. Independent assortment • The segregation of one pair of alleles does not affect the segregation of another pair. • There is a random arrangement of parental chromosomes at metaphase of meiosis.

  7. Monohybrid cross • Carry out a cross for a dominant and a recessive homozygote where P is for pink flower and p is for white • Cross the F1 and give the ratios of the F2.

  8. Test Cross • A cross carried out to determine an organism’s genotype, by mating it with a homozygous recessive organism. • Show how a test cross works

  9. Dihybrid Cross • Carry out a cross for a dominant and a recessive homozygote where Y is for yellow and y for green, and R for round and r for wrinkled. • Cross the F1 and give the ratios of the F2.

  10. Dominance Incomplete dominance Co-dominance Lethal Alleles

  11. Incomplete dominance • Action of one allele does not completely mask the action of the other. • Neither allele has dominant control over the trait. • Heterozygous offspring is intermediate in phenotype

  12. Eg snapdragons • P1: RR (Red) x rr (white) • F1: Rr (pink) • F2: ? (You determine the ratios) • 1red:2pink:1white

  13. Co-dominance • Both alleles in heterozygous organism contribute to the phenotype. • Both alleles are independently and equally expressed.

  14. Eg Human Blood Group AB • P1: AA (type A) x BB (type B) • F1: AB (type AB)

  15. Eg Coat colour in horses and cattle • P1: CRCR(red) x CrCr(white) • F1: CRCr(roan) • F2: ? You determine the ratios • 1 Red: 2 Roan: 1 white • Roan is a blend of both white hairs and red hairs

  16. Lethal Alleles • Mutations of a gene that produce a non-functional gene product and affect the organisms survival. • If dominant, may kill in single dose • If recessive, kills when homozygote

  17. Eg Manx cat • MM (normal tail) • MML (manx - no tail) • MLML (lethal – deformity of spine in embryo) • Carry out a cross for two heterozygotes. What is the phenotypic ratio?

  18. Eg Yellow mice • YY (lethal – yellow – terminates at blastocyst stage) • Yy (yellow) • yy (not yellow) • Again, what is the phenotypic ratio for a cross of heterozygotes?

  19. Eg Huntington’s disease • A dominant allele is lethal. • Nerve cell death in brain causing jerky involuntary movements and dementia. • Why does it persist in the human population? • Shows in adults 30-40 years

  20. Multiple alleles More than one allele possible at a gene locus

  21. There are three different alleles: A, B and O The alleles code for making the enzyme that hold the sugars together that make the different antigens on the RBC. Blood groups

  22. O is nonfunctioning (recessive) A is A antigen (dominant) B is B antigen (dominant) A and B antigens can act with other antibodies so must be matched for transfusion.

  23. Frequency in NZ

  24. Determine the blood types

  25. Dilemma • If a mother is type A and has a baby type B, can the father be type O? • Explain your answer. • You can now do the self check for this section.

  26. Gene-gene interactions When a characteristic is influenced by more than one gene at two different loci or even on different chromosomes altogether.

  27. Epistasis (standing upon) • Involves two non-allelic genes (different loci) • Action of one gene masks or alters expression of other genes • Three forms – collaboration, complementary, supplementary

  28. Eg Albinism • Occurs in rodent that are homozygous recessive for colour even if they have alleles for agouti or black fur. • The gene for colour is epistatic gene 1 gene 2 A  B  C coat colour show one colour/another colour

  29. Collaboration • Ratio 9:3:3:1 (Although the ratio is typical, it is unusual that some of the phenotypes may not have been shown in the parents) • Four different phenotypes depending on the presence or absence of certain genes

  30. P_R_ walnut P_rr pea ppR_ rose pprr single Carry out a cross for two heterozygotes – PpRr x PpRr Eg comb shape in chickens

  31. Supplementary Genes (Epistasis) • Ratio 9:3:4 • A dominant allele at one locus is necessary for the expression of alleles at another • Typically three phenotypes • Carry out a cross for two heterozygotes – CcBb x CcBb

  32. Eg Coat colour in mice • Gene C controls the production of melanin • Gene B indicates whether the colour is black or brown • Without the production of melanin, there will be no colour. gene C gene B enzyme 1 enzyme 2 no pigment  melanin produced  Black Brown C_B_ Black C_bb Brown cc__ No colour

  33. Complementary Genes • Ratio 9:7 • Development of a characteristic requires the presence of at least one dominant allele at both of 2 loci • Typically there are two phenotypes • Carry out a cross for two heterozygotes – PpQq x PpQq

  34. Eg Purple pigment in sweet pea flowers • Gene P makes white intermediate • Gene Q converts white to purple gene P gene Q enzyme 1 enzyme 2 Colourless  Colourless  Coloured precursorintermediateproduct (white pp_ _) (white P_ _ _) (purple P_Q_)

  35. Duplicate genes • Ratio 15:1 • A characteristic is developed if EITHER or BOTH of the dominant alleles at two loci is/are present. • Carry out a cross for two heterozygotes – AaBb x AaBb

  36. Eg Fruit width in Shepherd’s Purse • Gene A and B code for two different enzymes which can form wide fruit. • gene A gene B enzyme A enzyme B Substance  Active  Substance XProductY wide (A_B_,A_bb, aaB_) narrow (aabb)

  37. Practice • Self check page 111/112 • Create a table to compare • Exam questions • Study book pg 31 Qb

  38. A women who owned a purebred female albino (lacking pigments) poodle (an autosomal recessive phenotype) wanted white puppies, so she took the dog to a breeder, who said he would mate her female with an albino stud male, also from a pure stock. When six puppies were born they were all black, so the women sued the breeder, claiming that he replaced the stud with a black dog, giving her six unwanted puppies. You are called in as an expert witness, and the defence asks you if it is possible to produce black offspring from two pure-breeding recessive albino parents. • (a) Discuss what evidence you would give by explaining what gene-gene interrelationship is involved in each of the parents and using appropriate allele symbols, draw biochemicalpathways to obtain an albino phenotype and a black phenotype. • Clear well-labelled diagrams may be used to help you answer this question. • (b) Explain the expected possible F2 phenotypes ratios if two of the black puppies were allowed to interbreed.

  39. Exams 4U 2007 Q5 • It is definitely a form of epistasis (that is – there is obviously more than one gene involved. • We know of 3 types of epistasis; collaboration, supplementary, complementary • Use the process of elimination • There are only 2 phenotypes, therefore it is not collaboration • There is no intermediary product, therefore it is not supplementary • Both genes are required to create colour, therefore it is complementary

  40. So, both parent dogs are claimed to be pure breeding – therefore homozygous • What are our options for this with 2 genes? – • AABB, aabb, but also AAbb and aaBB • Note a P1 cross of AABB x aabb and a P1 cross of AAbb x aaBB both make the F1 generation AaBb, which produces the complementary ratio of 9:7 in the F2

  41. Pleiotropy • A single gene may produce a product that can influence a number of traits in the phenotype.

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